Question

The quantity y varies as the cube of (x+2) y=32 when x=0. Find y when x=1

Answer 1

Answer:

$y \: \alpha \: {(x + 2)}^{3} \\ y = k {(x + 2)}^{3} \\ k \: is \: a \: constant \\ when \: x = 0 \: \: y = 32 : \\ 32 = k {(0 + 2)}^{3} \\ 32 = k(8) \\ k = 4 \\ \therefore \: y = 4 {(x + 2)}^{3} \\ when \: x \: is \: 1 : \\ y = 4 {(1 + 2)}^{3} \\ y = 4 \times 27 \\ y = 108$